Sequence of continuous functions whose pointwise limit is discontinuous Give an example of a sequence of continuous functions $f_n:\mathbb{R} \to \mathbb{R}$ such that the pointwise limit $f:\mathbb{R} \to \mathbb{R}$ exists but is discontinuous.
My idea is $f_n(x)=\displaystyle\frac{x}{n}.$ Then the pointwise limit is $0$ but $f$ is discontinuous on $\mathbb{R}.$ Is that correct?
I'm having a hard time understanding the definition of pointwise convergence. I have this definition:
$\displaystyle\lim_{n\to \infty} f_n=f$ pointwise iff $\displaystyle\lim_{n\to \infty} f_n(x)=f(x)$
But then what is $f(x)$? Where does the $n$ go? In the example I gave, for instance,
$\displaystyle\lim_{n\to \infty} f_n(x)=\displaystyle\lim_{n\to \infty} \frac{x}{n}=0.$ But does this equal $f(x)$? Is $f(x)$=0 a discontinuous limit?
 A: The example you give has a point-wise limit of 0, which is not a discontinuous function.  Some standard examples of these kinds of functions are ones that kind of smoothly interpolate between two values, with the interpolating region getting smaller and smaller. For example, consider the sequence:
$$
f_n(x) = \tan^{-1}(nx)
$$
This is plotted for some various values of $n$ below, on the region $x\in [-1,1]$

Along with Suugaku's answer, this should give you the intuition you need about point-wise limits and continuity.
A: $f_n(x) = \left\{\begin{array}{ccc}
0 & \quad & x < 0\\
x^n & \quad & 0\leq x \leq 1\\
1 & \quad & x > 1
\end{array}\right.$. 
You can verify $f_n$ are continuous  for all $n\in\mathbb{N}$. However $x<0$ implies $f_n(x)\rightarrow 0$ as $n\rightarrow\infty$, $0\leq x<1$ implies $f_n(x)\rightarrow 0$ as $n\rightarrow\infty$ and $x\geq1$ implies $f_n(x)\rightarrow 1$ as $n\rightarrow\infty$ implying the pointwise limit is not continuous.
A: A piecewice linear example is:
$$f_n(x) = \left\{\begin{array}{ccc}
0 & \quad & x \le 0\\
nx & \quad & 0<x<\frac{1}{n}\\
1 & \quad & x\ge\frac{1}{n}
\end{array}\right.$$
A: Yes, $f(x) = 0$ according to what you wrote. However, your answer is not correct because $0$ is a continuous function. An example of what you want are the functions:
$$
f_n(x) = \arctan(n x),
$$
which converges pointwise to 
$$f(x) = \frac{\pi}{2}\mbox{sgn}(x) = \left\{\begin{array}{ccc}
-\pi/2 & \quad & x < 0\\
0 & \quad & x = 0\\
\pi/2 & \quad & x > 0
\end{array}\right.$$
a discontinuous function.
A: Each function $f_n$ in the sequence given by $f_n(x)=\frac{1}{x^2+\frac{1}{n}}$ is continuous on all of $\mathbb{R}$, but the pointwise limit of the sequence $f(x)=\frac{1}{x^2}$ is clearly discontinuous at $x=0$.
A: Consider the function $e^{-n\,.\,x^2}$. This function is continuous for all $n$, but limiting function will be discontinuous function.
